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A cup of tea cool from 80 degree Celsius to 60 degree celsius in one minute tab indent temperature is is 60 degree in cooling from 60 degree Celsius 50 degree it will take what is the answer options 1) 30 sec
2) 60sec
3) 90sec
4) 50sec

Datta Wakde , 5 Years ago
Grade 12
anser 2 Answers
Vikas TU

Last Activity: 5 Years ago

Rate of cooling ( dT/dt) = K[ (Ti+Tf)/2 - To] 
( Tf - Ti)/t = K[ ( 80 + 60)/2 - 60]
( 80-60)/1 = K[ 70 - 60]
20 = K× 10 
K = 20/10 = 2
in second condition, 
initial temperature ( Ti) = 60°C 
Final temperature ( Tf) = 50°C 
Time taken for cooling is t 
A/C Newton's law of cooling 
( 60 - 50)/t = 2 [ (60+50)/2 -60]
10/t = 2* 5 
10/t = 10  
t = 1 min.

Khimraj

Last Activity: 5 Years ago

According to Rate of cooling
dT/dt  = K((T_initial+T_final)/2 - T_surr) 
( 80-60)/1 = K(70-60)
K = 2
Time taken for cooling
(60 - 50)/t = 2 ((60+50)/2 -60)
t = 60 sec

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